Remove dude "D" from C2 leaving ABCEF. Both B and C can't occupy the center cabin because "F" would have to be next door in either C1 or C3, leaving AEF but "F" also can't be in C2 because B and C are in the two outside cabins making C2 = AE. That leaves BCDF unknown. Since B and C can't be paired, nor can C and D that pairs BD and CF. Since C1 is the only choice remaining for F they take C1 and BD take C3.

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